FIN HEAT TRANSFER CLASS NOTES FOR MECHANICAL ENGINEERING
FIN HEAT TRANSFER CLASS NOTES FOR
MECHANICAL ENGINEERING
Fin is a solid extended surface or a combined conduction convection system. Heat is transferred by Conduction within the solid and by convection from the fin surface in a PERPENDICULAR direction to that of conduction. Finally heat is lost to the surroundings. Fins increase the rate of heat transfer by increasing the surface area without increase of primary surface area. This increased surface area reduces the convective resistance and increases the rate of heat transfer economically. The fins should have large surface area per unit volume (>700 m^{2}/m^{3}) with small gas flow passages with natural laminar flow for increased rate of heat transfer. It increases the heat exchange between two fluids which are separated by a sold surface as in home radiators and cooling system in cars.
PURPOSE OF A FIN: To increase heat transfer by increasing the surface area without increase of primary surface area. FIN involves both (CONDUCTIVE + CONVECTIVE) HEAT TRANSFER. . Finally it loses heat to the atmosphere.
HOW FINS INCREASE HEAT TRANSFER: Convection equation q =h A dT Watts.
Undoubtedly heat transfer can be increased by increasing the
(i) temperature difference
(ii) the convection heat transfer coefficient
(iii) the contact surface area of the object
ΔT can’t be increased because of fin base temperature and environment temperature are normally fixed. Convective heat transfer coefficient can be increasing by using a fan to blow air over the surface. However , this has not been practically possible to increase ‘h’. Thus it is not feasible to change the first two options. Therefore, increasing the surface area seems to be the only choice possible. ‘This can be easily and economically achieved by fins.
TYPE OF FINS / SHAPES OF FINS
(i)Fins of constant area of cross section like rectangular, circular, square and elliptical fins
(ii) Fins with varying area of cross section like triangular, trapezoidal , parabolic and conical
(iii) Plate fins and wound fins from the two types above
Fig. Types of fins
PURPOSE OF DESIGNING A FIN
Fin should be most effective. It should be able to transfer many times the heat as compared without fins. It should also be cost effective.
Write the assumption used in the analysis of heat transfer through fins.
FIN PARAMETER
Fin parameter is represented by the letter ‘m’. It is given by the relation given below
m = (h P/k A_{c})^{0.5}
Where
h surrounding medium convective HT coefficient
P is the perimeter of the fin
k thermal conductivity of fin material
A_{c} cross sectional area of the fin if rectangular, A_{c} =b x t
Q. FIN EFFICIENCY FOR A SINGLE FIN
It is applicable for a single fin.
Ratio of the actual rate of heat transfer to the maximum rate of heat transfer through the fin. Maximum heat transfer occurs if the entire fin were at the base temperature.
The fin efficiency is η_{f }= actual q^{.}_{fin}/ q^{.}_{max} with base temperature all along the length
η_{f} = (PhA_{c}k)^{1/2}(T_{b}–T )/hPL (T_{b}–T )
= (kA_{c}/hP)^{1/2}(1/L)
=1/mL
Finally η_{f =}1/mL
Fin efficiency will always be less than one. This is
because assuming the temperature throughout
the fin is at the base temperature would increase
the heat transfer rate.
PRACTICAL APPLICATIONS OF FINS
(i) Electric motors
(ii) Car radiators
(iii) Pumps
(iv) Compressors
(v) Internal Combustion engines
(vi) Transformers
(vii) Condensers
(viii) Evaporators
(ix) Air craft engines
(x) Economizers of steam power plants
(xii) In nature, the ears of Jackrabbits and Fennec Foxes act as fins to release heat from within their bodies_{.}
Fin Performance Parameters
Description of Fin performance in three different ways
(i) Effectiveness of a single fin
(ii) Efficiency of a single fin
(iii)Overall efficiency of an array of fins
FIN EFFECTIVENESS OF A SINGLE FIN
It is applicable for a single fin. It is the ratio of heat transfer with fin to the rate to the heat transfer without fin. Thus it is greater than 1. Practically it should be at least greater than 2.3.
The formula for a fin of infinite length
Є_{f} = q._{ with fin}/ q._{without fin}= (P h A_{c }k)^{1/2}(T_{b}–T)/ h A_{c} (T_{b}–T)
= (Pk/h A_{c})^{1/2}
Є_{f} = (Pk/h A_{c})^{1/2}
Where A_{c }is the fin crosssectional area
For a rectangular fin, P = 2B + 2 t and A_{c} = B t
Where B is width of fin and t is the thickness of fin
Now if the fin is thin, Neglect 2t as compared to 2B
Therefore P/A_{c} = 2B/B t=2/t
Є_{f} = (2k/ht)^{1/2} = (1/Bi)^{0.5}
Biot number, Bi for a fin
Less value of Biot number is desirable for more effectiveness.
Biot number will be less if h is small i.e. gas environment all around the fin.
Further, Biot number will be less if thermal conductivity k of fin material is high.
Biot number will be less if the thickness of fin is less i.e. fin is thin.
ASSUMPTIONS USED IN FIN ANALYSIS

Onedimensional steady state heat conduction

Homogeneous and isotropic fin material, k is constant

No internal heat generation

Uniform crosssectional area

Uniform convective heat transfer coefficient h

Negligible contact thermal resistance

Negligible radiation

Always calculate Rate of heat transfer at the base of the fin
NOMENCLATURE OF FIN

Base of the fin or root of the fin: Where fin connects to the main body

Tip of the fin—The other free end of the fin

A_{c} is cross sectional area of the fin = b x t for a rectangular fin

A_{s }is the surface area of the fin= perimeter x length= (2b+2t) L
Fig. Details of a rectangular fin
DERIVATION OF BASIC DIFFERENTIAL EQUATION FOR FINS OF CONSTANT AREA OF CROSS SECTION
General partial equation for a fin is
d^{2}θ/dx^{2} –m^{2} θ=0
where parameter m = (hP/kA_{c})^{0.5}
θ is the temperature difference
DERIVATION
Using above assumptions, apply the Law of conservation of energy for a differential element of the fin. Consider two sections at distances x and x+ dx from the fin base. Note conduction takes place through the crosssectional area. Convection takes place on the surface area and perpendicular to the conduction direction.
conduction at distance ‘x’ = conduction at (x+dx) + convection through length dx
q_{x} = q_{x+dx} + dq_{conv}
q_{x+dx} =q_{x}+ (dq_{x}/dx) dx
From Fourier equation
Firstly q_{x}= –kA_{c} (dT/dx)
Secondly q_{x+dx} = –kA_{c} (dT/dx) + d(–kA_{c} (dT/dx) /dx) dx
Simplify q_{x+dx} = –kA_{c} (dT/dx) — kA_{c}d(dT/dx) /dx) dx
q_{x+dx} = –kA_{c} (dT/dx) — kA_{c} d^{2}T/dx^{2}dx
dq_{conv} = h dA_{s} (T_{x} –T_{∞})
dA_{s }= Pdx is the surface area of the differential element
Where P is fin perimeter = (2b+2t) for a rectangular fin
Substituting
0 = k A_{c} (d^{2}T/dx^{2} )dx + h Pdx (T_{x} –T_{∞})
Since A_{c} is constant, it becomes_{ }
d^{2}T/dx^{2} =(hP/kA_{c}) θ where (T_{x} –T_{∞}) =θ & d^{2}T/dx^{2} =d^{2}θ/dx^{2 }^{ }
d^{2}θ/dx^{2} –m^{2} θ=0
Where fin parameter m = (hP/kA_{c})^{0.5}
It is a second order differential equation.
θ is the temperature difference
The solution will be
θ(x) = C_{1}e^{mx} + C_{2}e^{mx} = C_{3}cosh mx+C_{4 }Sinh mx
Both exponential and hyperbolic solutions are equivalent.
Find the constants C_{1}, C_{2}, C_{3} & C_{4} from the boundary conditions.
h surrounding medium convective HT coefficient
K thermal conductivity of fin material
A_{c} cross sectional area of the fin if rectangular=b x t
Finally the equation is
d^{2}θ/dx^{2} –m^{2} θ=0
It is a second order differential equation.
Where fin parameter m = (hP/kA_{c})^{0.5}
ϴ is the temperature difference θ_{x} = T_{x}— T_{a}
The solution of the differential equation will be
ϴ(x) = C_{1}e^{mx} + C_{2}e^{mx} = C_{3}cosh mx+C_{4 }sinh mx
Both exponential and hyperbolic solutions are equivalent solutions.
This is a common equation for the three cases.
Find the constants C_{1}& C_{2} (or C_{3} and C4)from the boundary conditions.
h surrounding medium convective HT coefficient
k thermal conductivity of fin material
A_{c} cross sectional area of the fin if rectangular, A_{c} =b x t
THREE DIFFERENT CASES OF FIN ANALYSIS
Fig. Different cases of fin
Case 1: Infinitely long fin: To find Rate of heat transfer and temperature distribution for a very long fin
Boundary conditions (a) at x=0, ϴ=ϴ_{0 }= T_{b}T_{a}
(b) At x = ∞, T =T_{a}_{, }and thus ϴ= T_{a}T_{a} =0
(1) Put x=0, C_{1} +C_{2} = ϴ_{0}
(ii) Put x =∞, we get C_{1} e^{∞}+ C_{2}e^{∞} =0 = C_{1} e^{∞}
therefore C_{1} =0 and then C_{2} = ϴ_{0}
_{ ϴ (x) =ϴ0 emx }
_{ϴ(x)/ ϴ0 = emx}
(T_{x} –T_{a})/( T_{b} –T_{a}) = e^{–mx}
At distance x from the fin base, Find the temperature. Thus the temperature distribution becomes known.
Rate of heat transfer at the base of the fin will be
q^{.} =(hPkA_{c})^{0.5}( T_{b} –T_{a})
Case2: Insulated fin tip: To find Rate of heat transfer and temperature distribution for insulated fin tip
The boundary conditions will be
(1) at x=0, ϴ=ϴ_{0 }= T_{b}T_{a}
(2) at x =L, dϴ/dx=0
Temperature distribution
(T_{x} –T_{a})/( T_{b} –T_{a}) =Cosh m(Lx)/Cosh mL
Rate of heat transfer at the base of the fin will be
q^{.} =(hPkA_{c})^{0.5}( T_{b} –T_{a}) Tanh mL
Case 3: Real finite fin: To find Rate of heat transfer and temperature distribution for a fin of finite length:
Boundary conditions are
(1) at x=0, ϴ=ϴ_{0 }= T_{b}T_{a}
(2) at x=L, the rate of heat transfer by convection on area A_{c }is equal to the rate of conduction through the area A_{c}
Temperature distribution
(T_{x} –T_{a})/( T_{b} –T_{a}) =
[Cosh m(Lx) + (h/km)Sinh m(L—x)]/[Cosh mL + (h/km) Sinh mL]
Rate of heat transfer at the base
q^{.} =(h P k A_{c})^{0.5}( T_{b} –T_{a}) Tanh mL [Tanh mL + h/km]/ [ 1 + (h/km) Tanh mL]
From the above three cases , case I and case II are much simpler than case III. Fins in actual use will correspond to case III.
When to use analysis of case I and case II for case III?
(i) Use equations of case I for case III when mL=5 with maximum error in calculations has been found to be < 1 % which is negligible.
(ii) Use equations of case II for case III, using corrected length in place of L in case II
Corrected length = actual length + 1/2 thickness of fin
With corrected length, maximum error in calculations has been found to be < 1 % which is negligible.
Fig. Heat loss in 3 cases of fins
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